A 9.780-g gaseous mixture contains ethane (C2 H6) and propane (C3 H8) . Complete combustion to f

A 9.780-g gaseous mixture contains ethane (C2 H6) and...

A $9.780-\mathrm{g}$ gaseous mixture contains ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ and propane $\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .$ Complete combustion to form carbon dioxide and water requires 1.120 mole of oxygen gas. Calculate the mass percent of ethane in the original mixture.

Key Concepts

Mass Percent Composition

Mass percent composition expresses the proportion of each component in a mixture as a percentage of the total mass. This concept is key in analytical chemistry for understanding the makeup of a mixture, requiring the calculation of the mass of individual components relative to the total mass.

System of Equations in Mixture Analysis

When dealing with mixtures, particularly in combustion problems, it is often necessary to set up and solve simultaneous equations that represent mass balances and stoichiometric relationships. This process is crucial for determining the individual amounts of each component in a mixture based on known total quantities and reaction stoichiometry.

Combustion Reactions

This concept involves chemical reactions in which a fuel reacts with oxygen to produce combustion products such as carbon dioxide and water. It is fundamental in understanding energy release and chemical changes, and it requires a clear identification of reactants and products along with balanced equations.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It uses balanced chemical equations to determine the amounts of substances consumed and produced, allowing for calculations based on mole ratios, mass, and volume of reactants or products.

Transcript

00:01 We have a mixture of ethylene and then we're going to burn it in oxygen.

00:05 So the total mass of the mixture is 9 .780 gram, and then we have 1 .12 more of oxygen being consumed.

00:14 So what would be the mass percentage of ethan in the mixture? okay, so first of all i will convert everything, all the known mass to number of modes.

00:30 However, we don't know the number, the mass of individual of both components.

00:37 So that's why we're going to assume that we will x gram of the e -thing, where we want to find the mass percentage of e -thing.

00:46 Okay, so we are going to convert to number of more by using x.

00:53 So for a number more eithing, or c2h6, is equal to x, divided by the number of more, motor mass.

01:02 The motor mass of ethane is equals to 30.

01:09 Okay, so we will have the number of more of ethene, and then we would like to know the number of more oxygen being consumed by the combustion of the thing.

01:20 So we just, you can see that the motor ratio is 1 to 7 over 2, or 1 to 3 .5.

01:28 So we will have 3 .5 times x, divide by and this is the number of more oxygen required for eithing.

01:39 So this is for you think.

01:42 All right, so and then for propane, we're going to do this same coalition.

01:48 Because the to full mixture is 9 .780, so we have the overall mass subtract eggs.

01:55 We should have the mass of propane.

01:57 So, and then we're going to divide it by the mass of propane.

02:02 It will be equal to 44.

02:04 All right, so the number of more oxygen is, required for propane combustion will be equal to the number of most time 5 because the ratio is 1 to 5...

Please give Ace some feedback